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Solving for Oxidation

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Hi, I need assistance with the following two questions:

1)Determine the oxidation numbers for all elements in the reaction below:

6NaOH(aq) + 2Al(s) ----> 3Hsub2(g) + 2Nasub3AlOsub3(aq)

b) Given 2.0g of solid Al in a cleaner, how much hydrogen gas will be produced?

2) Write the balanced molecular equation, complete the ionic equation and net ionic equation for the reaction of strontium nitrate(soluble) with potassium iodate(soluble) to form strontium iodate(precipitate) and potassium nitrate(soluble).

Solution Summary

This solution provides a detailed explanation of how to solve the above chemistry-based problems, and for required calculations, gives a step by step guide of how to derive the answer. After using this solution, a student should be able to understand this subject material much better. Tips regarding what to remember when solving these kinds of problems are provided.

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1) Determine the oxidation numbers for all elements in the reaction:
6NaOH(aq) + 2Al(s) ----> 3H2(g) + 2Na3AlO3(aq)

The oxidation number of an element in a simple binary ionic compound is the number of electrons gained or lost by an atom of that element when it forms the compound. Therefore, in NaOH, sodium is +1 and OH- is -1. Therefore, oxygen is -2 and hydrogen is +1.

Aluminum's oxidation number is zero since it's in the elemental form. The same goes for H2, the elemental form of hydrogen.

For Na3AlO3, the sodium ions are the same as before. They're +1. Combined, AlO3 is -3, the actual charge on the ion, since the ...

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